Analytical Applications unit test
A test on this unit alone, marked as a percentage and a letter grade — for the test your class is actually sitting, rather than for May. Answer everything, then submit once: seeing the answer to question 3 before attempting question 4 makes the final percentage meaningless.
Sketching f from f′
Optimization: write the constraint
Reading f′ to describe f
Because the graph turns around
f′(3) = 0 justifies nothing
Sign information gives shape, not height
Comparing f at two points using f′
Candidates Test procedure
Reading a graph of f′ to answer about f
Report both coordinates
Local versus absolute
Justifying an absolute extremum
Short answer 1. Define or explain: MVT gives an open-interval c
3 ptsShort answer 2. Define or explain: Optimization: justify the extremum
3 ptsShort answer 3. Define or explain: Ties are allowed
3 ptsShort answer 4. Define or explain: Second derivative test
3 ptsFree response
9 ptsNO CALCULATOR. Let f be a twice-differentiable function on the closed interval [−4, 4] with f(2) = 3. The graph of f′, the derivative of f, is shown: it starts at (−4, −0.5), dips to a local minimum at (−3, −1), rises to cross the x-axis at x = −2, continues up to a local maximum at (1, 3), falls through (2, 1.5) to touch the x-axis at x = 3 (where f′(3) = 0), and then rises steeply to (4, 5). So f′ < 0 on (−4, −2), f′ > 0 on (−2, 3) and (3, 4], and f′ = 0 at x = −2 and x = 3.
A. For x > 0, the function g is defined by g(x) = f(x) − ln x. Find g′(2). Show the work that leads to your answer.
B. Find all values of x on the open interval 0 < x < 3 at which the graph of f has a point of inflection. Give a reason for your answer.
C. For −4 ≤ x ≤ 4, on what open intervals, if any, is the graph of f both increasing and concave down? Give a reason for your answer.
D. For −4 ≤ x ≤ 4, find the value of x at which f has an absolute minimum and the value of x at which f has an absolute maximum. Give reasons for your answers.